To show that a relation is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.

1. Reflexive: A relation R is reflexive if for every a in N≥0, (a, a) is in R. In this case, (b - a) would be (a - a) which equals 0. Since 0 is divisible by 6, the relation is reflexive.

2. Symmetric: A relation R is symmetric if for every (a, b) in R, (b, a) is also in R. In this case, if (b - a) is divisible by 6, then -(a - b) is also divisible by 6. Therefore, the relation is symmetric.

3. Transitive: A relation R is transitive if for every (a, b) in R and (b, c) in R, (a, c) is also in R. In this case, if (b - a) and (c - b) are both divisible by 6, then their sum (c - a) = (b - a) + (c - b) is also divisible by 6. Therefore, the relation is transitive.

Therefore, the relation R is an equivalence relation.

Question

To show that a relation is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.

1. Reflexive: A relation R is reflexive if for every a in N≥0, (a, a) is in R. In this case, (b - a) would be (a - a) which equals 0. Since 0 is divisible by 6, the relation is reflexive.

2. Symmetric: A relation R is symmetric if for every (a, b) in R, (b, a) is also in R. In this case, if (b - a) is divisible by 6, then -(a - b) is also divisible by 6. Therefore, the relation is symmetric.

3. Transitive: A relation R is transitive if for every (a, b) in R and (b, c) in R, (a, c) is also in R. In this case, if (b - a) and (c - b) are both divisible by 6, then their sum (c - a) = (b - a) + (c - b) is also divisible by 6. Therefore, the relation is transitive.

Therefore, the relation R is an equivalence relation.

Knowee AI · Accepted Answer

To show that a relation is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.

1. Reflexive: A relation R is reflexive if for every a in N≥0, (a, a) is in R. In this case, (b - a) would be (a - a) which equals 0. Since 0 is divisible by 6, the relation is reflexive.

2. Symmetric: A relation R is symmetric if for every (a, b) in R, (b, a) is also in R. In this case, if (b - a) is divisible by 6, then -(a - b) is also divisible by 6. Therefore, the relation is symmetric.

3. Transitive: A relation R is transitive if for every (a, b) in R and (b, c) in R, (a, c) is also in R. In this case, if (b - a) and (c - b) are both divisible by 6, then their sum (c - a) = (b - a) + (c - b) is also divisible by 6. Therefore, the relation is transitive.

Therefore, the relation R is an equivalence relation.

Let R be a relation on the set N≥0 given byR = {(a, b) : (b − a) is divisible by 6}Show that this is an equivalence relation

Question

Let R be a relation on the set $N_{\geq 0}$ given by $R = \{(a, b) : (b - a) \text{ is divisible by } 6\}$ Show that this is an equivalence relation.

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